Skip to content
Related Articles

Related Articles

Total sum except adjacent of a given node in BST
  • Difficulty Level : Easy
  • Last Updated : 04 Nov, 2020
GeeksforGeeks - Summer Carnival Banner

Given a BST and a key Node, find the total sum in BST, except those Node which are adjacent to key Node. 

Examples:

1:-First find the total sum of BST 
2:-Search the key Node and trace its parent Node. 
3:-If the key Node is present then, subtract the sum of its adjacent Node from total sum 
4:-If key is not present in BST then return -1.

C++




// C++ program to find total sum except a given Node in BST
#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int data;
    struct Node *left, *right;
};
 
// insertion of Node in Tree
Node* getNode(int n)
{
    struct Node* root = new Node;
    root->data = n;
    root->left = NULL;
    root->right = NULL;
    return root;
}
 
// total sum of bst
int sum(struct Node* root)
{
    if (root == NULL)
        return 0;
 
    return root->data + sum(root->left) + sum(root->right);
}
 
// sum of all element except those which are adjecent to key Node
int adjSum(Node* root, int key)
{
 
    int parent = root->data;
 
    while (root != NULL) {
        if (key < root->data) {
            parent = root->data;
            root = root->left;
        }
        else if (root->data == key) // key Node matches
        {
            // if the left Node and right Node of key is
            // not null then add all adjecent Node and
            // substract from totalSum
            if (root->left != NULL && root->right != NULL)
                return (parent + root->left->data +
                                 root->right->data);
 
            // if key is leaf
            if (root->left == NULL && root->right == NULL)
                return parent;
 
            // If only left child is null
            if (root->left == NULL)
                return (parent + root->right->data);
 
            // If only right child is NULL
            if (root->right == NULL)
                return (parent + root->left->data);
        }
 
        else {
            parent = root->data;
            root = root->right;
        }
    }
 
    return 0;
}
 
int findTotalExceptKey(Node *root, int key)
{
    return sum(root) - adjSum(root, key);
}
 
// Driver code
int main()
{
    struct Node* root = getNode(15);
    root->left = getNode(13);
    root->left->left = getNode(12);
    root->left->left->left = getNode(11);
    root->left->right = getNode(14);
    root->right = getNode(20);
    root->right->left = getNode(18);
    root->right->right = getNode(24);
    root->right->right->left = getNode(23);
    root->right->right->right = getNode(25);
    int key = 20;
    printf("%d ", findTotalExceptKey(root, key));
    return 0;
}

Java




// Java program to find total sum
// except a given Node in BST
class GFG
{
static class Node
{
    int data;
    Node left, right;
};
 
// insertion of Node in Tree
static Node getNode(int n)
{
    Node root = new Node();
    root.data = n;
    root.left = null;
    root.right = null;
    return root;
}
 
// total sum of bst
static int sum(Node root)
{
    if (root == null)
        return 0;
 
    return root.data + sum(root.left) +
                       sum(root.right);
}
 
// sum of all element except those
// which are adjecent to key Node
static int adjSum(Node root, int key)
{
    int parent = root.data;
 
    while (root != null)
    {
        if (key < root.data)
        {
            parent = root.data;
            root = root.left;
        }
        else if (root.data == key) // key Node matches
        {
            // if the left Node and right Node of key is
            // not null then add all adjecent Node and
            // substract from totalSum
            if (root.left != null && root.right != null)
                return (parent + root.left.data +
                                root.right.data);
 
            // if key is leaf
            if (root.left == null && root.right == null)
                return parent;
 
            // If only left child is null
            if (root.left == null)
                return (parent + root.right.data);
 
            // If only right child is null
            if (root.right == null)
                return (parent + root.left.data);
        }
        else
        {
            parent = root.data;
            root = root.right;
        }
    }
    return 0;
}
 
static int findTotalExceptKey(Node root, int key)
{
    return sum(root) - adjSum(root, key);
}
 
// Driver code
public static void main(String[] args)
{
    Node root = getNode(15);
    root.left = getNode(13);
    root.left.left = getNode(12);
    root.left.left.left = getNode(11);
    root.left.right = getNode(14);
    root.right = getNode(20);
    root.right.left = getNode(18);
    root.right.right = getNode(24);
    root.right.right.left = getNode(23);
    root.right.right.right = getNode(25);
    int key = 20;
    System.out.printf("%d ",
               findTotalExceptKey(root, key));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program to find total sum
# except a given Node in BST
class getNode:
     
    def __init__(self, n):
         
        self.data = n
        self.left = None
        self.right = None
 
# Total sum of bst
def sum(root):
     
    if (root == None):
        return 0
 
    return (root.data + sum(root.left) +
                        sum(root.right))
 
# Sum of all element except those
# which are adjecent to key Node
def adjSum(root, key):
     
    parent = root.data
 
    while (root != None):
        if (key < root.data):
            parent = root.data
            root = root.left
        elif (root.data == key):
             
            # Key Node matches
            # if the left Node and right Node of key is
            # not None then add all adjecent Node and
            # substract from totalSum
            if (root.left != None and
               root.right != None):
                return (parent + root.left.data +
                                 root.right.data)
 
            # If key is leaf
            if (root.left == None and
               root.right == None):
                return parent
 
            # If only left child is None
            if (root.left == None):
                return (parent + root.right.data)
 
            # If only right child is None
            if (root.right == None):
                return (parent + root.left.data)
 
        else:
            parent = root.data
            root = root.right
 
    return 0
 
def findTotalExceptKey(root, key):
     
    return sum(root) - adjSum(root, key)
 
# Driver code
if __name__ == '__main__':
     
    root = getNode(15)
    root.left = getNode(13)
    root.left.left = getNode(12)
    root.left.left.left = getNode(11)
    root.left.right = getNode(14)
    root.right = getNode(20)
    root.right.left = getNode(18)
    root.right.right = getNode(24)
    root.right.right.left = getNode(23)
    root.right.right.right = getNode(25)
     
    key = 20
     
    print(findTotalExceptKey(root, key))
     
# This code is contributed by bgangwar59

C#




// C# program to find total sum
// except a given Node in BST
using System;
 
class GFG
{
class Node
{
    public int data;
    public Node left, right;
};
 
// insertion of Node in Tree
static Node getNode(int n)
{
    Node root = new Node();
    root.data = n;
    root.left = null;
    root.right = null;
    return root;
}
 
// total sum of bst
static int sum(Node root)
{
    if (root == null)
        return 0;
 
    return root.data + sum(root.left) +
                       sum(root.right);
}
 
// sum of all element except those
// which are adjecent to key Node
static int adjSum(Node root, int key)
{
    int parent = root.data;
 
    while (root != null)
    {
        if (key < root.data)
        {
            parent = root.data;
            root = root.left;
        }
        else if (root.data == key) // key Node matches
        {
            // if the left Node and right Node of key is
            // not null then add all adjecent Node and
            // substract from totalSum
            if (root.left != null && root.right != null)
                return (parent + root.left.data +
                                 root.right.data);
 
            // if key is leaf
            if (root.left == null && root.right == null)
                return parent;
 
            // If only left child is null
            if (root.left == null)
                return (parent + root.right.data);
 
            // If only right child is null
            if (root.right == null)
                return (parent + root.left.data);
        }
        else
        {
            parent = root.data;
            root = root.right;
        }
    }
    return 0;
}
 
static int findTotalExceptKey(Node root, int key)
{
    return sum(root) - adjSum(root, key);
}
 
// Driver code
public static void Main(String[] args)
{
    Node root = getNode(15);
    root.left = getNode(13);
    root.left.left = getNode(12);
    root.left.left.left = getNode(11);
    root.left.right = getNode(14);
    root.right = getNode(20);
    root.right.left = getNode(18);
    root.right.right = getNode(24);
    root.right.right.left = getNode(23);
    root.right.right.right = getNode(25);
    int key = 20;
    Console.Write("{0} ",
            findTotalExceptKey(root, key));
}
}
 
// This code is contributed by PrinciRaj1992
Output: 
118





 

Time Complexity: O(n) + O(h) where n is number of nodes in BST and h is height of BST. We can write time complexity as O(n).
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up
Recommended Articles
Page :